Question

Question 18
Show that p – 1 is a factor of $p^{10}–1$ and also of $p^{11}–1$.

Answer 1

Let $g\left(p\right)=p^{10}–1$ ... (i)
And $h\left(p\right)=p^{11}–1$ …(ii)
On Putting P = 1 in Eq. (i) we get,
$g\left(1\right)=1^{10}–1=1–1=0$
Hence, p – 1 is a factor of g(p) [Using Remainder Theorem]
Again, putting p = 1 in Eq (ii) we get
$h\left(1\right)=(1{)}^{11}–1=1–1=0$
Hence, p – 1 is a factor of h(p) [Using Remainder Theorem].